Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $q = \dfrac{n^2 - 10n + 9}{n - 1} \div \dfrac{-8n + 72}{2n - 8} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{n^2 - 10n + 9}{n - 1} \times \dfrac{2n - 8}{-8n + 72} $ First factor the quadratic. $q = \dfrac{(n - 9)(n - 1)}{n - 1} \times \dfrac{2n - 8}{-8n + 72} $ Then factor out any other terms. $q = \dfrac{(n - 9)(n - 1)}{n - 1} \times \dfrac{2(n - 4)}{-8(n - 9)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (n - 9)(n - 1) \times 2(n - 4) } { (n - 1) \times -8(n - 9) } $ $q = \dfrac{ 2(n - 9)(n - 1)(n - 4)}{ -8(n - 1)(n - 9)} $ Notice that $(n - 1)$ and $(n - 9)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ 2\cancel{(n - 9)}(n - 1)(n - 4)}{ -8(n - 1)\cancel{(n - 9)}} $ We are dividing by $n - 9$ , so $n - 9 \neq 0$ Therefore, $n \neq 9$ $q = \dfrac{ 2\cancel{(n - 9)}\cancel{(n - 1)}(n - 4)}{ -8\cancel{(n - 1)}\cancel{(n - 9)}} $ We are dividing by $n - 1$ , so $n - 1 \neq 0$ Therefore, $n \neq 1$ $q = \dfrac{2(n - 4)}{-8} $ $q = \dfrac{-(n - 4)}{4} ; \space n \neq 9 ; \space n \neq 1 $